Animated Solution for Chemistry - d and f-Block Elements: If ${\text{Fe}}^{3+}$ and ${\text{Cr}}^{3+}$ both are present in group III of qualitative analysis, then distinction can be made by

The world of qualitative analysis is like a grand chemical puzzle. When you are handed a test tube containing a mixture of unknown ions, your job is to act as a molecular detective. You must use the principles of solubility, common ion effect, and redox chemistry to systematically separate and identify each culprit.

In this classic problem, we are faced with two transition metal heavyweights: Iron (${\text{Fe}}^{3+}$) and Chromium (${\text{Cr}}^{3+}$). Both belong to Group III of the qualitative analysis scheme. Let's dive into the elegant chemistry that allows us to tell them apart.

A Delicate Balance

The first step in our investigation is to isolate these ions from the rest of the mixture. We do this by exploiting their solubility products. Group III cations precipitate as hydroxides, but there is a catch: we only want Group III to precipitate, not the cations of subsequent groups like Magnesium or Calcium.

To achieve this delicate balance, we use a buffer system of ammonium hydroxide (${\text{NH}}_4\text{OH}$) and ammonium chloride (${\text{NH}}_4\text{Cl}$). The ammonium chloride provides a high concentration of ammonium ions (${\text{NH}}_4^{+}$), which suppresses the ionization of the weak base ammonium hydroxide due to the common ion effect.

This ensures that the concentration of hydroxide ions (${\text{OH}}^{-}$) is just high enough to exceed the solubility product ($K_{sp}$) of the highly insoluble ${\text{Fe(OH)}}_3$ and ${\text{Cr(OH)}}_3$, but too low to precipitate the hydroxides of Group IV and V.

When we add this reagent mixture, a beautiful but messy precipitate forms. We get a mixture of reddish-brown iron(III) hydroxide and green chromium(III) hydroxide.

Oxidation is Key

Now we have a mixed precipitate. How do we separate them? We can't just dissolve them in acid, because both will dissolve. We need a chemical scalpel—a reaction that affects one but ignores the other.

This is where redox chemistry comes to the rescue. We treat the mixed precipitate with sodium hydroxide ($\text{NaOH}$) and bromine water (${\text{Br}}_2$).

When bromine reacts with sodium hydroxide, it undergoes a disproportionation reaction to form sodium hypobromite ($\text{NaBrO}$), which is a powerful oxidizing agent. It readily releases highly reactive nascent oxygen:

Here is where the intrinsic properties of the transition metals shine.

Iron is already in its $+3$ oxidation state. In an aqueous alkaline medium, this is its highest stable oxidation state. It looks at the powerful oxidizing agent and simply shrugs. The ${\text{Fe(OH)}}_3$ remains completely unaffected and stays as an insoluble reddish-brown solid.

Chromium, on the other hand, has a trick up its sleeve. While it is currently in the $+3$ state, it can be oxidized to the $+6$ state. The nascent oxygen aggressively attacks the chromium hydroxide, oxidizing it to form sodium chromate (${\text{Na}}_2{\text{CrO}}_4$).

The magic of sodium chromate is that it is highly soluble in water. As the reaction proceeds, the green precipitate vanishes, and the solution turns a brilliant, clear yellow.

We have achieved separation! We simply filter the mixture. The iron stays behind on the filter paper as a solid, while the chromium passes through into the filtrate as a yellow solution.

Painting it Yellow

To put the final nail in the coffin and absolutely confirm the presence of chromium, we take that yellow filtrate and acidify it slightly with acetic acid. Then, we add a few drops of lead acetate (${\text{Pb(CH}}_3{\text{COO)}}_2$).

Instantly, a heavy, bright yellow precipitate of lead chromate (${\text{PbCrO}}_4$) crashes out of the solution.

This classic "yellow paint" precipitate is the definitive proof that chromium was present in our original mixture. Through a masterful combination of solubility control and selective oxidation, we have successfully solved the puzzle.